GATE CSE 1993 | Question: 01.1

Question

The eigen vector $(s)$ of the matrix $$\begin{bmatrix} 0 &0 &\alpha\\ 0 &0 &0\\ 0 &0 &0 \end{bmatrix},\alpha \neq 0$$ is (are)

• A. $(0,0,\alpha)$
• B. $(\alpha,0,0)$
• C. $(0,0,1)$
• D. $(0,\alpha,0)$

Comments

Ans

---

x3 should be 0 ;       x1, x2 can be anything  from the above ques , 
see this example by @Ayush Upadhyaya

---

@goku4199 The question is not ambiguous. You are searching for $(1,0,0)^T$ and $(0,1,0)^T$ as options
is incorrect because these are just linearly independent vectors. The question
asks what eigenvectors are possible, and $\alpha$ is just a constant multiplier.
Hence, options B and D are the correct answers.

Answer 1

Since, the given matrix is an upper triangular one, all eigenvalues are 0. And hence $A-\lambda I=A$

So the question asks

$\begin{bmatrix} 0 &0 &\alpha \\ 0&0 &0 \\ 0& 0& 0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

what $x_1,x_2,x_3$ are suitable?

which means $x_1$times column1+$x_2$times column2+$x_3$ times column3=Zero Vector

Look carefully,if you take any linear combination of columns 1 and 2 and don't take column 3(means $x_3=0$), the result will be a zero vector.

So, if we have our X=$\begin{bmatrix} \beta\\ \gamma\\ 0 \end{bmatrix}$ where $\beta, \gamma$ are any scalar quantity,

but $x_3$ must be necessarily zero to get zero vector.

Hence, only option (B) and (D) satisfy.

Comments

Himanshu3 No, zero vector can’t be eigen vector of a matrix.

https://math.stackexchange.com/questions/990016/can-the-zero-vector-be-an-eigenvector-for-a-matrix

---

I don't understand why option D is correct can anyone explain me in easy way?

---

@halfcodeblood just follow definition $Av= \lambda v$   As $\lambda=0$  So Given Eigen vectors After Multiply With matrix $A$ must be $0$ so check again now i hope u'll get it.

Answer 2

Given matrix is an upper triangular matrix. Therefore diagonal elements of A are eigen values of A, i.e. λ=0,0,0.

Consider the characteristic equation $(A-λI)x=0$

$\Rightarrow$ $\begin{bmatrix} 0-\lambda & 0 & \alpha \\ 0 & 0-\lambda & 0\\ 0 & 0 & 0-\lambda \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

To find the eigen vector of A corresponding to $\lambda$
Put the eigen value of A in the characteristic equation

i.e. $\begin{bmatrix} 0 & 0 & \alpha \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$ $\begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}$ = $\begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix}$

Now we have

$\alpha$$x_3$=0

$\Rightarrow$$x_3$ = 0

Therefore, any non zero vector with z-component as zero is an eigen vector.
Hence in the given options $B$ and $D$ are correct.

Comments

Thank you for explaining :)

---

This explanation was better

Answer 3

Answer: $B, D$

Eigen values are: $0,0,0$

The eigen vector should satisfy the equation: $\alpha$z $=$ $0$

Comments

if B & D are true then A should also be true please check once?

---

Ok

---

You have to think in this way too as we know Ax= λx . here λ =0 then Ax=0, and when Ax will give Zero whenever it multiplies with zero . so we can put only those value non zero for which they are making ultimately zero. but As we can see in option first if u multiply with this then it will  become  α^2 . which doesnt satisy the condtion coz α is not equal to zero here.

Answer 4

 

 

solution

Answer 5

Guys this is an easy question, so let us not make it complex.

As most of the people have already explained that $|A-\lambda I|*eigenvector =\begin{bmatrix} 0 &0 &\alpha \\ 0 & 0 &0 \\ 0& 0 & 0 \end{bmatrix}$$\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$

Now see that we are getting $0*x_{1} + 0*x_{2} + \alpha *x_{3}=0$ as the only useful equation.

Now put whatever the value of $x_{1}$ ,$x_{2}$ and $x_{3}$, we will get our eigenvectors, EXCEPT the cases where $\alpha$ has to be zero because in question it clearly mentioned that $\alpha \neq 0$. This condition will only fail when $x_{3}\neq 0.$ And therefore now you can match the options very easily.

Comments

shukriya bhai i was banging my head on this for 15 minutes

 

Answer 6

clearly matrix is upper triangular matrix and for upper triangular ,lower triangular matrix ,diagonal matrix and scaler matrix ,its diagonal element is eigen values.so here eigen value=0,0,0.

now we have to find eigen vector of this matrix (A-λ)X=0  where λ=0 so we get A matrix  [ 0 0 α] has rank=1  means we get two free variable (n-r) 

x1= β x2=γ x3=0 then eigen vector  (x1,x2,x3)=(β,γ,0)= β(1,0,0)+γ(0,1,0) so  B and  D  are  right answer .